Lecture9C(groundconvergencederivation)

8/13/2019 Lecture9C(groundconvergencederivation)

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d r d r r r 111 )()(

1sideonForceRadial

==

d r d r r r 333 )()(

3sideonForceRadial==

)2

()()2

sin()()(

2sideonForceNormal

2312

d

dr d

r r ==

)

2

()()

2

sin()()(

4sideonForce Normal

4314

d

dr d

r r ==

[ ]dr r 2)(

4sideonForceShear

=

[ ]dr r 4)(2sideonForcehear


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direction)tangential(involumeunit per componentforce body

direction)radial(involumeunit per componentforce body==

S

R

directionradialinforceEquation of Equilibrium

( ) ( )[ ] 02

)(2

)()()(

42

4231

=+

+

dr Rrd dr

d dr

d dr d r d r

r r

r r

Divide both sides by drd

[ ] ( ) ( )

0)()(21)()( 42

42

31

=+

++

Rr d dr r r r r r r

Area gets smallerto the limit

d

r r

r r

)(


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[ ] ( ) ( ) 0)()(21)()( 42

4231 =+++ Rr

d dr r r r r r r

0=++ Rr d

r

r r r r r

r r

r r

r r

r r

+

=

+

=

)()(

+

r r

r r

Expand by product rule

Divide the equation by r

01 =++

+

R

r r r

r r r

Similarly the equation of equilibrium in the tangential direction

021 =++

+

S

r r r

r r


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Differential equation of equilibrium in Polar Coordinate Differential equation of equilibrium in Cartesian Coordinate

0=+

+

X y x xy x

01

=+

+

+

Rr r r r r r

021 =++

+

S

r r r

r r

0=+

+

Y

x y

xy y

Make use of Airy stress function to find the elastic stress distribution of a circular hole in aninfinite medium under isotropic stress condition

o

i P

P

PressureExternal

PressureInternal


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Differential equation of equilibrium in Cartesian Coordinate

0=+

+

X

y x

xy x 0=+

+

Y

x y

xy y

g =For no horizontal acceleration, X=0 Body force is simply the weight of the body

0=++ g

x y xy y 0=

+

y x xy x

Also need the Compatibility Equations in terms of Stresses

xu

x =

Differentiate twice w.r.t. y

Components of Strain

yv

y =

xv

yu

xy +

=

Differentiate twice w.r.t. x Differentiate w.r.t. x and then y

Arrive the Compatibility Equation in terms of Strain

y x x y xy y x

=

+

2

2

2

2

2


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Use Hooks Law to transform into Compatibility Equation in terms of Stress

For Plane Stress Condition :

)(1 y x x E = )(1 x y y E

= xy xy xy E G )1(21 +==

Substitute into the compatibility equation, we get:

y x x y xy x y y x

+=

+

2

2

2

2

2

)1(2)()(

Next, to differentiate the equilibrium equation

0=

+

y x

xy x

0=+

+

g x y

xy y

Differentiate w.r.t. x

Differentiate w.r.t. y and then add thetwo equations and it becomes:

2

2

2

22

2 y x y x

y x xy

=


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The Compatibility Equation in terms of Stresses is obtained by substituting

back into

y x x y xy x y y x

+=

+

2

2

2

2

2

)1(2)()(

2

2

2

22

2 y x y x y x xy

=

)1()()(

2

2

2

2

+=

+

x y

x y y x )( 2

2

2

2

y x

y x

)1(22

2

2

2

2

2

2

+=

+

x x y y x y y x )( 2

2

2

2

y x

y x

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

y x y x x x y y y x y x x y y x

=

+


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02

2

2

2

2

2

2

2

=

+

+

+

y x x y y x y x

( ) 022

2

2

=+

+

y x y x The Compatibility Equation in terms of Stresses

For Plane Stress Condition :

The Compatibility Equation in terms of Stresses for Plane Stress Conditionincluding body forces are similarly derived and given as follows:

( ) ( )

+

+=+

+

yY

x X

y x y x 12

2

2

2


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For Plane Strain Condition :

( ) ( )[ ] y x x E += 111 2

( ) ( )[ ] x y y E += 111 2

xy xy xy E G )1(21 +==

( ) 022

2

2

=+

+

y x y x The Compatibility Equation in terms of Stresses

For Plane Stress Condition also the same for

Plane Strain Condition


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The Compatibility Equation in terms of Stresses for any general case with body forces are similarly derived as follows:

( )

+

=+

+

yY

x X

y x y x

11

2

2

2

2

The usual method of solving both the equilibrium and compatibilityequations is to introduce a new stress function (introduced byAiry in 1862 , that is why call Airy Stress Function)

0=

+ y x xy x

0=+

+

g x y

xy y

For the equilibrium equations

to be satisfied, it has been shown that the new stress function must also satisfy the following expressions for the stresscomponents :

gy

x y

=2

2

gy y x

=

2

2

y x xy =

2


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y x xy =

2Substitute these expressions

into the equilibrium equation

gy y x

=

2

2

gy x y

=

2

2

( ) 022

2

2

=+

+

y x y x

The stress function must also satisfy the following expressions:

024

4

22

4

4

4

=

+

+

y y x x

If this stress function can be satisfied, then the equilibriumequation and compatibility equation will also be satisfied .

Thus, the solution of many 2D problems (including bodyforces) can be derived by finding a solution which satisfy thestress function incorporating the boundary conditions.


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The stress function (without body force) in polar coordinatesmust also satisfy the following expressions for the stresscomponents:

=

=

r r r r r r 111 2

22

2

2

11

+

=

r r r r 22

r =

The Compatibility Equation in terms of Stresses in polar coordinates is:

01111

2

2

22

2

2

2

22

2

=

+

+

+

+

r r r r r r r r


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Stress Distribution Symmetrical about an Axis

The Compatibility Equation in terms of Stresses in polar coordinates, when the Stress Function depends on r only, is:

01111

2

2

22

2

2

2

22

2

=

+

+

+

+

r r r r r r r r

011211

32

2

23

3

4

4

2

2

2

2

=++=

+

+dr d

r dr d

r dr d

r dr d

dr d

r dr d

dr d

r dr d

Which is an ordinary differential equation , which can be reduced to a linear differential equation with constantcoefficients by introducing a new variable t such that

The solution has four constants of integration, which must be determined from the boundary conditions.

t

er =


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Assume the stress function to be in the form as:

The corresponding stress function without body force is obtained:

DCr r Br r A +++= 22 loglog

2

2

2

11

+

= r r r r C r Br A

r r r 2)log21(1

2 +++=

=

C r Br A

2)log23(2 +++= 22

r

=

=

r r r

10= r

The solution becomes

which may be used to represent the stress

distribution in a hollow cylinder subjected touniform pressure on the inner and outer surfaces

C r A

r 22 += C r A

22 += 0= B For


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( ) ( ) obr r iar r p p == == Apply the boundary conditions:

C r A

r 22 +=

ir pC a

A

=+= 22 or pC b A

=+= 22 From which,

( )22

22

ab

p pba A io

=22

22

2 ab

b pa p

C oi

=

C r A

22 += C r A

r 22 += and substitute back into the equation


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( )22

22

222

22 1ab

a pa pr ab

p pba oiior

+

=

( )22

22

222

22 1ab

a pa pr ab

p pba oiio+

=

r uor r u ==

r E =

To find the radial displacement, use

For Plane Stress Condition


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0112

32

2

23

3

4

4

=++dr d

r dr d

r dr d

r dr d

check

DCr r Br r A +++= 22 loglog DCr r A ++= 2log

Cr r

Adr d

21 += C

r A

dr d

21

22

2

+= 44

4 16

r A

dr d =

33

3 12

r A

dr d =

4

16

r A 3

12

2r

Ar

+ )21(1 22 C r A

r + 0)21(13 =++ Cr r

Ar

C r r

A 211

24 + 0121 24 =++ r

C r

A41

6r

A 41

4r

A+

016 4 =+ r A416 r A


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Derivation of Elasto-Plastic Solution to compute radial crown displacement of a tunnel under planestrain, homogeneous, isotropic stress condition for c and material

er i p o pa

o p


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For soil under elastic stresses, use the equations of equilibrium and compatibility for a cylindricalhole in an infinite plate

01 =++ r r r rr r rr

Equations of equilibrium

without body force

021

=+

+

r r r r r

Equations of equilibrium

without body force

( ) 011 22

22

2

=+

+

+

rr

r r r r

Equations of compatibility

without body force

( ) 022

2

2

=+

+

y x y x

( ) +=+ r y x


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0=

For no shear stress in the tangential directions

0=+ r r rr rr

( ) 01

2

2

=+

+

rr r r r

r r

k rr Ar =+

( ) 0122

=+

+

rr r r r

( ) ( ) 222

2

2

)1( ==+ k k

rr r k Ak Ar r r ( ) ( ) 21111 ===+

k k k rr Akr Akr r

Ar r r r r

Assumes that the equation takes the form of:

and substitute into

0=+ rr rr

2


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( ) 0122

=+

+

rr r r r

0)(0)(

22

22

==+

k Ar k k k Ar

k

k + 2)1( k r k Ak 02 =k Akr

thereforeandzero bemust,zero becannotsince k A

Arr =+

0=+

r r

rr rr Substitute back into the equilibrium equation

( ) 0=+

r

Ar r

rr rr rr

02 =+

r A

r r rr rr


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For a linear, first order differential equation, the solution is given as follows:

For a linear, first order differential equation, which is expressed in the following form:

( ) ( ) xQ y x P dxdy =+

+= cdxQe ye Pdx Pdx

r A

r r rr rr =+

2Analogy yrr

( ) x P r 2

( ) xQr A

The solution by integration is from standard mathematics handbook:

+= C dr er A

edr

r dr

r rr

22

dxr

+= C dr er A

e r r rr )ln(2)ln(2


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+= C dr er A

e r r rr )ln(2)ln(2

+= C Ardr r rr 2

2)ln(2 r e r =

C r

Ar rr += 2

22

22 r C A

rr +=

Arr

=+

Ar

C A

=++ 22

22 r C A =

rerr e

orr

r r at pr

====

,, +=

C A po 2 o p A 2=

2e

or r

C pe +=

The two boundary conditions: at

( ) 2rpC =


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( )( ) 2er o

eor

r pC

r pC

e

e

=

=

22 r C A

rr += ( )2

=r r

p p er oorr e

( )2

+=r r p p er oo e

r

uu

r r +

=

1

r u r =

( )( ) 01

=+= rr zz zz E ( ) += rr zz

which is the radialstress in the elasticzone

22 r C A =

The equations for expressing tangential strain is given as:

For zero shear strain:

For plane strain condition:

In order to express displacement due to stress changes plus initial stresses


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In order to express displacement due to stress changes plus initial stresses

)(o

p+

)( orr rr p+

( )( ) zz rr E += 1 ( ) += rr zz

( )( )[ ] ++= rr rr E 1

( ) ( ) ( ) ( )( )( )[ ]oorr orr o p p p p E ++= 1

( )( ) ( )( )[ ]orr o p p E += 111 2

r u r

=

r u r

( )( ) ( )( )[ ]orr o p p E += 111 2

+1


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( )( ) ( )[ ]orr or p p E r u += 1

1

( )2

=r r

p p er oorr e

( )2

+=r r

p p er oo e

( )2

1

+=r r p

E r u e

r or

e which is the elastic radial displacement

For the stresses in the plastic zone, needs to use the Mohr-Coulomb failure criterion

cohesiontheiscwhere N c N rr 2+=

cohesionresidual theis

cwhere N c N r r rr r r

2+=r

r

r

N

sin1

sin1

+=

For Peak Strength

For Residual Strength

NN 2+=


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0=+

r r

rr rr Substitute r r rr

N c N r

2+= into

02

=++

r

N c N

r r r r rr rr rr

( )r

N c

r

N

r r r r rr rr

21 =

( ) ( ) xQ y x P dxdy =+

+= cdxQe ye Pdx Pdx

+=

C dr er

N cedr

r

N

r

dr r

N

rr

r

r

r 11

12

( ) ( ) += C dr er

N ce r N

r r N

rr r

r

r ln1ln1 12

1


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+= C dr r N cr r r r N r N rr

1

2)1(

+= C dr r N cr r

r r

N r N

rr

2)1(

C r N

N cr

r

r

r r

N r N

rr ++= +

1

)1( 112

( ) ( ) C r N N cr r r

r r N r N rr += 1)1( 1

12

1

1

r r x x dx C

r

+

= ++

( ) C r N

N cr

r

r

r r

N r N

rr ++=

1

)1( 112

par == To solve for C use the boundary conditions at


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( ) ( ) C r N N cr r r

r r N r N rr

+= 1)1( 11

2

irr par ,

( ) ( ) C a N N ca p r r

r r N r N i += 1)1( 1

12

( ) ( )1)1( 112

+=r

r

r r N r N i

a N N c

a pC

To solve for C, use the boundary conditions at

Substitute C into

( ) ( ) C r N N cr r r r r N r N rr +

=

1)1( 1

12


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( ) ( ) +

= 1)1(1

12

r

r

r r N r N rr

r N

N c

r

( ) ( )11

1

12

+r

r

r r N r N i

a N

N c

a

p

( )12

)1(

2)1(

+

=

r

r

r

r r

N

N c

N

N c

par r r

i

N

rr

which is the radial stresses

in the plastic zone

r

r r N

sin1

sin1+

=( )( ) ( ) ( )

r

r

r r

r

r r

r r

r

r r

N

sin1cos

sin1)sin1(sin1

sin1)sin1(sin1)sin1(

sin1sin1 2

=

=

+=

+=

r

r

r

r r

r

r r

N

sin1sin2

sin1sin1)sin1(

1sin1sin1

1

=

++=+=

tN 22)1( NNN


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( ) 2cot

sin2cos

1r

r

r

r

r

N

N

== and substitute back into

to obtain the radial stress in the plastic zone

( )12

)1(

2)1(

+

=

r

r

r

r r

N

N c

N

N c p

ar r r

i

N

rr

[ ] r r r r i N

rr cc par r

cotcot

)1(

+

=

which is the radial stressesin the plastic zone

To locate the extent of the plastic zone , rewrite

as follows: N c N rr 2+=

cohesion peak iscand angle friction peak iswhere

crr

sin1cos

2sin1sin1

+

+=

Take r = r e2


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Substitute into

( )2

=r r

p p er oorr e ( )2

+=r r

p p er oo e

sin1cos

2sin1sin1

+

+= crr

( )ee r

e

er oorr r

r p p =

=

2

( ) ( ee r oo

e

er oo p pr

r p p +=

+=

2

( ) sin1 cos2sin1 sin1 ++=+ c p p ee r r oo

sin1

cos2

sin1

sin12

+

+= c p

ee r r o

ee r r o c p

sin1sin1

sin1sin1

sin1cos

2+

+=

2cos r


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sin1sin1cos

2

=

er o c p

( )

( )

cossin1

sin1sin1

cossin1

c p

c p

or

or

e

e

=

=

which is the radial stress atthe elastic-plastic boundary

[ ] r r r r i N

err cc pa

r r

cotcot)1(

+

=

( ) cossin1 c por e ==

Determine the plastic radius by substituting the above equation back into

[ ]

cos)sin1(cotcot)1(

c pcc par

or r r r i

N e

r

=+

[ ]

cos)sin1(cotcot)1(

cpccpr N e

r

=+


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[ ] cos)sin1(cotcot c pcc pa or r r r i

+

2

)1(

cotcotcos)sin1(

X c p

cc par

r r i

r r o N

er

=+

+=

1)1( X N r =

1

cotcotcos)sin1(

ln

+

+

= r r r i

r r o

N

c pcc p

e aer

2

1

X ar

X

e

=

21 lnln X ar

X e

=

1

2ln X X

e aer =Plastic zone exist only when r e is greater than the radius a, thus:

1

2ln X X

aea = 1ln

1

2

=e X X


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aea

1

0ln

0ln

1

2

2

1

2

=

==

X

X

X X

e

1

cot

cotcos)sin1(2 ==

+

+ X

c p

cc p

r r i

r r o

r r ir r o c pcc p crit cotcotcos)sin1( +=+

cos)sin1( c p p oicrit = which is the critical internalsupport pressure

For the plastic zone to form,crit ii p p p


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Next is to find out the displacement in the plastic zone

r u r =

strainradial r u

rr r

=

Strain is consisted of two parts: elastic and plastic E P

P rr

E rr

r

r u +=

angledilationtheiswhere N

sin1sin1

+

=

Plasticity Rule is based on 3 assumptions:

1 Th i ld it i (F) t d b f i th t


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1. The yield criterion (F), represented by a surface in the stress space,at which plastic deformation may develop

2. The hardening law (h being a hardening parameter), that governs the possible changes in shape, size and position of the yield surface withan increase in plastic strains

3. The plastic flow rule governing the increment of the plastic strains

For elasto-plastic materials undergoing infinitesimal deformation,Total strain increment = Elastic strain increment + Plastic strain increment

pl el d d d += Plastic strains are irreversibleThe elastic strain component can be represented using the generalized Hooke's law:

pl el d d D Dd d == D = Elastic Constitutive Modulus

1st assumption in Plasticity Rule:


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The yield criterion (F), represented by a surface in the stress space, atwhich plastic deformation may develop

pl h F F ,=

h is the vector of the hardening parameters governing the changes of the yield surface withincreasing plastic strains

If F < 0 , material is elastic (stress state within the yield surface)

If F = 0, material is in plastic equilibrium (stress state fulfills the yield criterion or stressstate at the yield surface)

F > 0 Not Admissible (stress state cannot be outside the yield surface)

2nd and 3 rd assumptions in Plasticity Rule:


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2. The hardening law (h being a hardening parameter), that governs the possible changes in shape, size and position of the yield surface withan increase in plastic strains

3. The plastic flow rule governing the increment of the plastic strains

Plastic Flow Rule states that during the plastic strain increment along theyield surface, the plastic strain increment is proportional to the gradient ofthe Plastic Potential

Plastic Potential pl hQQ ,=

= Q

d d pl

d Plastic Multiplier IncrementThis means that the vector representing the plastic strain increment is directed as the

outward normal vector at the point that corresponds to the current stress state.

Plastic Flow Rule states that during the plastic strain increment along theyield surface, the plastic strain increment is proportional to the gradient of


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the Plastic Potential

Plastic Potential pl

hQQ ,=

= Qd d pl

d Plastic Multiplier IncrementThis means that the vector representing the plastic strain increment is directed as the

outward normal vector at the point that corresponds to the current stress state.

The plastic potential is defined by Johnson and Mellor (1983) as follows:


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( ) ( ) 0sin =+= r r g

sin1sin1

+= N

( ) 0sinsin == r r g ( ) ( ) 0sin1sin1 =+= r g

( )( ) ( )( ) 0sin1 sin1sin1 sin1 =+=

r g

( )( ) 0sin1

sin1

=+

=

r g

0== r N g

Using the flow rule, the plastic strain increments may be determined as:


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multiplier scalar negativenonais g

d ij

P ij

= ,

For the radial and tangential strain increments:

( )( )

( )( ) 0sin1

sin1sin1

sin1=

+

=

r g [ ] sin1 =

P r d

sin1 =

r

g

sin1 = g

[ ] sin1 = P d

( )

N

d

d P

P

r =

+=

sin1sin1

N

P

P

r =

P rr

E rr

r

ru +=Substitute into

N r =

P

P


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r

P E rr

r N r u =

P E += E P =

( ) E E rr r N r

u =

=

E r E

rr r

r u

N r

u

E E rr r

r N ur

N

r u

+=+

( ) ( ) xQ y x P dxdy =+ E E

rr r r N u

r N

ru

+=+


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r r

+= cdxQe ye Pdx Pdx

( ) ++= dr e N C eu dr r N

E E rr

dr r

N

r

++= dr N r dr r C r u E N E

rr N N

r

++= dr N r dr r C r u E N E rr

N N r

For plane strain conditions, recall the tangential strain

( )( )

( )( )[ ]orr o

p p E

+=

111 2 E

( ) ( )( ) ( )[ ]orr o E p p E

+= 11

Substitute the residual strength parameters r r

N c N r rr 2+=


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into

( )( )( ) ( )[ ]orr o

E

p p E +

=

11

( )( ) [ ]( ) ( )[ ]orr or rr E p p N c N E r r ++= 21

1

( )( ) ( ) ( )( ) ( )[ ]orr or orr E p N p N c p N E r r r +++= 121

1

( )( ) ( ) ( )( ) ( )[ ]orr or orr E p N p N c p N E r r r +++= 1211

( ) ( ) ( ) ( )( ) ( ) ++

+=12

121

r r r

r r r

N p N c p N

p N p N c p N

E or orr

orr or orr E

( ) ( ) ( ) ( )+++ 121 pNpNcpN


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( ) ( ) ( ) ( )

( ) ( )

+=12

121

r r r

r r r

N p N c p N

p N p N c p N

E or orr

orr or orr E

( )( )( ) ( ) ( )( )[ 1211 +++= r r r r N p N c p N N E or orr E

( )( ) ( )( )[ 1+ E [ ] r r r r i

N

rr cc par r

cotcot

)1(

+

=

( ) ( ) 121 ++=r r r r

N p N c p N N E or orr

( ) ( )( )

[ ]

( ) ( )( ) ++

+

+=

121

cotcot1

1

r r

r

r r

N p N c

pcc par

N N

E or

or r r r i

N

E

For plane strain conditions, recall the radial strain and derive same manner as tangential strain

E 1


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47

( )( ) zz rr E rr E

+=1

( ) += rr zz ( )( )[ ] ++= rr rr E rr E 1

( ) ( ) ( ) ( )( )( )[ ]oorr oorr E rr p p p p E ++= 1

( )( ) ( )( )[ ]oorr E

rr p p E += 111 2

( )( ) ( )[ ]oorr E

rr p p E +

=

11

Substitute the residual strength parameters r r

N c N r rr 2+=


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48

into

( )( ) ( )[ ]oorr E

rr p p E +

=

11

( )( ) ( ) ( )( )[ ]r r r N c N p p N p E r oorr orr E

rr

2111

+++

=( )( ) ( )[ ]1211 +=

r r r N p N c N p

E or orr E

rr

Substitute the radial stress equation in the plastic zone

into the above equation

[ ] r r r r i N

rr cc par r

cotcot)1(

+

=

( )( ) ( )[ ]or rr orr E rr p N c N p E r r ++= 211

( )( ) ( )[ ]1211 += r r r N p N c N p E or orr E

rr


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[ ] r r r r i N

rr cc par r

cotcot

)1(

+

=

( )( ) ( )+

+=

121cotcot1

1

r r r

r

N p N c N pcc par

E or or r r r i

N E

rr

Substitute into the radial displacement equation for the plastic zone (radial and tangential strain just derived)

++= dr N r dr r C r u E N E rr N N r

( ) ( )( )

[ ]

( ) ( )( ) ++

+

+=

121

cotcot1

1

r r

r

r r

N p N c

pcc par

N N

E or

or r r r i

N

E

dr r For E rr N


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50

( )

( ) ( ) dr N p N c N pcc p

a

r

E r

r r r

r

or or r r r i

N N

+

+

121cotcot1

1

( ) ( )( )( ) ( )( )[ ] +++

++=

+

dr r N p N c N pc

dr r a

c p N

E N or or r

N N

N r r i

r r r

r

r r

121cot

cot11

1

1

( ) ( )

( )( ) ( )( )[ ]++++

++

= +

+

+

121cot1

cot1

11

1

r r r

r r

r

r

N p N c N pc N r

a

c p N

N N r

E or or r

N

N r r i

N N

dr r For E N


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( ) ( )

( )

[ ]

( ) ( )( )

dr

N p N c

pcc par N N

E r

r r

r

r r

or

or r r r i

N

N

++

+ +

121

cotcot1

1

( ) ( ) ( ){ }( ) ( ) ( )( )[ ] +++

+++=

+dr r N p N c N N pc

dr r

a

c p N N

E N or or r

N N

N r r i

r r r r

r

r r r

121cot

cot1

1

1

( )( ) ( )

{ }( ) ( ) ( )( )+++++

++++=

++

+

1211cot1

cot

111

1

r r r r

r r r

r

r

N p N c N r

N N pc N r

ac p N N

N N r

E or

N

or r

N

N r r i N N

Assign the following parameters for simplicity in expressions


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52

( )1cot

+= r N

r r i

l a

c p B

r

r

N N

r r H

N N

+

+

=)(

1)(

1

+

=

+

N

r r H

N

( ) ( )+

+ cot1 1

r c pN

r N

r r i N N

Substitute


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53

( ) ( )

( )( ) ( )( )[ ]+++++

= +

+

121cot1

11

1

1

r r r

r r

r

N p N c N pc N r

a N

N N

E or or r

N

N

( )( ) ( )

{ }( ) ( ) ( )( )+++++

++++=

++

+

1211

cot1

cot

111

1

r r r r

r r r

r

r

N p N c N r N N pc

N r

a

c p N N

N N r

E or

N

or r

N

N r r i

N N

and

into

++= dr N r dr r C r u E N E rr N N r

[ ]21 )()( l r H l r H C r u N r ++=


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54

( )( ) ( )[ ] l B N N N N E l r r r ++= 111

( ) ( )( ) ( )( ) ( ) ( )( )( )[ or r or pc N N N N N N p N c E

l r r r r r

++++

=

cot1112

12

To solve for C, use the continuity condition at r=r e

( )2

1 += r u er ( ) ( )rpru +=1


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55

( )

=r

p

E r

er o

r e

( ) ( ) er oer r p E r u e

( ) [ ]21 )()( l r H l r H C r r u ee N eer ++=

( ) =+ er o r p E e 1 [ ]21 )()( l r H l r H C r ee N e ++

( ) 211 )()(1 l r H l r H r p E

C ee N

er o e+= +

Documents

Lecture9C(groundconvergencederivation)